AS & A Level Physics (9702)
Ideal gases Banker
30 questions · about 3% of the marks we indexed · examined in 9 of 9 sessions.
Practice questions
Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not
Cambridge past paper questions and have not been reviewed by Cambridge. Try each one before opening the answer.
Q1Kinetic theory of gases · Core · Paper 4 style · 7 marks- (a) State four assumptions of the kinetic theory of gases. [4]
- (b) Explain, in terms of molecules, how a gas exerts a pressure on the walls of its container. [3]
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- (a) Any four of: molecules are in continuous random motion; collisions (with each other and the walls) are perfectly elastic; the volume of the molecules is negligible compared with the volume of the gas; there are no intermolecular forces (except during collisions); the time of a collision is negligible compared with the time between collisions; there is a very large number of molecules.
B1 each for any four of the listed assumptions
- (b) Molecules collide with the wall and rebound, so their momentum changes; the wall exerts a force on each molecule and so (Newton's third law) the molecules exert a force on the wall; many collisions per second give a steady average force, and pressure = force/area.
B1 change of momentum on collision with wall; B1 force = rate of change of momentum (and Newton's third law); B1 pressure = total force/area of wall
Air in a room is at 27 °C (300 K). Nitrogen molecules have a mass of 28 u. Take the Boltzmann constant k = 1.38 × 10⁻²³ J K⁻¹ and 1 u = 1.66 × 10⁻²⁷ kg.
- (a) Calculate the mean translational kinetic energy of a nitrogen molecule. [2]
- (b) Calculate the root-mean-square speed of the nitrogen molecules. [2]
- (c) State how the mean kinetic energy of an oxygen molecule in the same air compares with that of a nitrogen molecule. [1]
Show answer
- (a) 6.21 × 10⁻²¹ J
⟨Eₖ⟩ = (3/2)kT (C1) = 1.5 × 1.38 × 10⁻²³ × 300 = 6.21 × 10⁻²¹ J (A1)
- (b) 517 m s⁻¹
m = 28 × 1.66 × 10⁻²⁷ = 4.65 × 10⁻²⁶ kg; ½m⟨c²⟩ = 6.21 × 10⁻²¹ (C1); c_rms = √(2 × 6.21 × 10⁻²¹/4.65 × 10⁻²⁶) = 517 m s⁻¹ (A1)
- (c) It is the same, because the mean kinetic energy depends only on the temperature.
B1
A sealed flask contains an ideal gas. Take the Boltzmann constant k = 1.38 × 10⁻²³ J K⁻¹ and 1 u = 1.66 × 10⁻²⁷ kg.
- (a) Calculate the mean translational kinetic energy of the molecules at 100 °C. [2]
- (b) A student says that heating the gas from 0 °C to 100 °C makes the mean kinetic energy of the molecules 'infinitely larger'. Calculate the actual ratio (mean KE at 100 °C)/(mean KE at 0 °C) and explain the student's error. [2]
Show answer
- (a) 7.72 × 10⁻²¹ J
T = 373 K (C1); ⟨Eₖ⟩ = 1.5 × 1.38 × 10⁻²³ × 373 = 7.72 × 10⁻²¹ J (A1)
- (b) ratio = 373/273 = 1.37; the mean KE is proportional to the thermodynamic (kelvin) temperature, not to the Celsius temperature.
B1 1.37; B1 must use kelvin: ⟨Eₖ⟩ ∝ T in K, and 0 °C is not zero energy
- (a) Show that the equation pV = ⅓Nm⟨c²⟩ can be written as p = ⅓ρ⟨c²⟩, where ρ is the density of the gas. [1]
- (b) Air at sea level in Gwadar has a pressure of 1.01 × 10⁵ Pa and a density of 1.20 kg m⁻³. Calculate the root-mean-square speed of the air molecules. [2]
- (c) Suggest why individual air molecules travel much less than 500 m in one second along a straight line. [1]
Show answer
- (a) Nm is the total mass of the gas and Nm/V = ρ, so p = ⅓(Nm/V)⟨c²⟩ = ⅓ρ⟨c²⟩.
B1
- (b) 502 m s⁻¹
⟨c²⟩ = 3p/ρ = 3 × 1.01 × 10⁵/1.20 = 2.53 × 10⁵ m² s⁻² (C1); c_rms = 502 m s⁻¹ (A1)
- (c) They collide very frequently with other molecules, changing direction (random path).
B1
- (a) Explain, in terms of the motion of molecules, why the pressure of a fixed mass of gas increases when it is heated at constant volume. [3]
- (b) A sealed can of air at 1.0 × 10⁵ Pa and 20 °C is left on a car dashboard in Hyderabad in summer, where it reaches 60 °C. Calculate the new pressure. Assume the volume of the can is constant. [2]
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- (a) Higher temperature → greater mean kinetic energy, so molecules move faster; each collision with the walls gives a larger change in momentum and collisions happen more often; so a larger force per unit area (pressure).
B1 mean KE/speed of molecules increases; B1 larger momentum change per collision; B1 more frequent collisions → greater rate of change of momentum → greater pressure
- (b) 1.14 × 10⁵ Pa
p/T constant, T in kelvin: 293 K → 333 K (C1); p = 1.0 × 10⁵ × 333/293 = 1.14 × 10⁵ Pa (A1)
A container of volume 0.040 m³ holds 2.4 × 10²⁴ molecules of nitrogen, each of mass 4.65 × 10⁻²⁶ kg. The pressure of the gas is 2.5 × 10⁵ Pa.
- (a) Explain what is meant by the root-mean-square speed of the molecules of a gas. [1]
- (b) Calculate the mean-square speed ⟨c²⟩ of the molecules. [2]
- (c) Hence calculate the r.m.s. speed. [1]
Show answer
- (a) The square root of the mean of the squares of the speeds of all the molecules.
B1
- (b) 2.69 × 10⁵ m² s⁻²
⟨c²⟩ = 3pV/(Nm) (C1) = 3 × 2.5 × 10⁵ × 0.040/(2.4 × 10²⁴ × 4.65 × 10⁻²⁶) = 2.69 × 10⁵ m² s⁻² (A1)
- (c) 518 m s⁻¹
√2.69 × 10⁵ = 518 m s⁻¹ (B1)
Take the Boltzmann constant k = 1.38 × 10⁻²³ J K⁻¹ and 1 u = 1.66 × 10⁻²⁷ kg.
- (a) Use the equations pV = ⅓Nm⟨c²⟩ and pV = NkT to show that the mean translational kinetic energy of a molecule is (3/2)kT. [3]
- (b) Calculate the temperature at which the mean translational kinetic energy of a helium atom is 1.0 × 10⁻²⁰ J. [2]
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- (a) ⅓Nm⟨c²⟩ = NkT, so m⟨c²⟩ = 3kT and ½m⟨c²⟩ = (3/2)kT.
M1 equate the two expressions; M1 m⟨c²⟩ = 3kT; A1 multiply by ½ to obtain ⟨Eₖ⟩ = ½m⟨c²⟩ = (3/2)kT
- (b) 483 K
T = 2⟨Eₖ⟩/(3k) (C1) = 2 × 1.0 × 10⁻²⁰/(3 × 1.38 × 10⁻²³) = 483 K (A1)
In a computer model of a gas, five molecules have speeds of 300 m s⁻¹, 400 m s⁻¹, 500 m s⁻¹, 600 m s⁻¹ and 700 m s⁻¹.
- (a) Calculate the mean speed of the five molecules. [1]
- (b) Calculate the root-mean-square speed of the five molecules. [2]
- (c) Explain why the r.m.s. speed is always greater than the mean speed unless all the speeds are equal. [1]
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- (a) 500 m s⁻¹
B1 2500/5 = 500 m s⁻¹
- (b) 520 m s⁻¹
⟨c²⟩ = (9 + 16 + 25 + 36 + 49) × 10⁴/5 = 2.70 × 10⁵ m² s⁻² (C1); c_rms = √(2.70 × 10⁵) = 520 m s⁻¹ (A1)
- (c) Squaring gives extra weight to the larger speeds, so the square root of the mean of the squares exceeds the mean.
B1
Take the Boltzmann constant k = 1.38 × 10⁻²³ J K⁻¹ and 1 u = 1.66 × 10⁻²⁷ kg.
- (a) Describe the shape of a graph of c_rms (y-axis) against thermodynamic temperature T (x-axis) for an ideal gas. [2]
- (b) The r.m.s. speed of the molecules of a gas at 300 K is v. State the temperature at which the r.m.s. speed is 2v. [1]
- (c) Calculate the temperature at which the r.m.s. speed of carbon dioxide molecules (mass 44 u) is 400 m s⁻¹. [2]
Show answer
- (a) A curve through the origin with decreasing gradient: c_rms ∝ √T.
B1 passes through the origin; B1 curve with gradient decreasing (c_rms ∝ √T)
- (b) 1200 K
B1 c_rms ∝ √T so T increases by a factor of 4
- (c) 282 K
T = m⟨c²⟩/(3k) (C1) = 44 × 1.66 × 10⁻²⁷ × 400²/(3 × 1.38 × 10⁻²³) = 282 K (A1)
Argon is a monatomic gas. Take R = 8.31 J K⁻¹ mol⁻¹, N_A = 6.02 × 10²³ mol⁻¹ and k = 1.38 × 10⁻²³ J K⁻¹.
- (a) Explain why, for an ideal monatomic gas, the internal energy is equal to the total translational kinetic energy of its atoms. [2]
- (b) Show that the internal energy U of n moles of an ideal monatomic gas at thermodynamic temperature T is (3/2)nRT. [2]
- (c) Calculate the increase in internal energy when 2.0 mol of argon is heated from 300 K to 400 K. [2]
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- (a) There are no intermolecular forces in an ideal gas, so the potential energy of the atoms is zero; a monatomic gas has only translational kinetic energy.
B1 no intermolecular forces so no potential energy; B1 internal energy = sum of KE (and PE) of the particles
- (b) U = N × (3/2)kT, and Nk = nN_A × (R/N_A) = nR, so U = (3/2)nRT.
M1 U = N(3/2)kT; A1 Nk = nR
- (c) 2490 J
ΔU = (3/2)nRΔT (C1) = 1.5 × 2.0 × 8.31 × 100 = 2490 J (A1)
A computer simulation gives the mean-square speed ⟨c²⟩ of the atoms of a monatomic gas at different thermodynamic temperatures T.T/K: 250, 300, 350, 400, 450⟨c²⟩/10⁵ m² s⁻²: 1.56, 1.87, 2.18, 2.49, 2.81It is suggested that ⟨c²⟩ = 3kT/m, where m is the mass of one atom. k = 1.38 × 10⁻²³ J K⁻¹ and 1 u = 1.66 × 10⁻²⁷ kg.
- (a) Explain why a graph of ⟨c²⟩ against T should be a straight line through the origin, and state an expression for its gradient. [2]
- (b) Determine the gradient of the graph. [2]
- (c) Use your gradient to determine the mass of one gas atom, in kg and in u. [2]
- (d) Predict the value of c_rms at 600 K. [1]
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- (a) ⟨c²⟩ = (3k/m)T, which has the form y = mx with no intercept; gradient = 3k/m.
B1 ⟨c²⟩ ∝ T (y = mx); B1 gradient = 3k/m
- (b) 625 m² s⁻² K⁻¹
M1 large triangle, e.g. (450, 2.81 × 10⁵) and (250, 1.56 × 10⁵); A1 gradient = 1.25 × 10⁵/200 = 625 m² s⁻² K⁻¹ (allow 620–630)
- (c) 6.62 × 10⁻²⁶ kg = 40 u (argon)
C1 m = 3k/gradient = 3 × 1.38 × 10⁻²³/625; A1 6.62 × 10⁻²⁶ kg, which is 40 u
- (d) 612 m s⁻¹
B1 ⟨c²⟩ = 625 × 600 = 3.75 × 10⁵ m² s⁻², c_rms = 612 m s⁻¹
- (a) Calculate the temperature at which the mean translational kinetic energy of the particles of a gas is 1.0 eV. (1 eV = 1.60 × 10⁻¹⁹ J, k = 1.38 × 10⁻²³ J K⁻¹) [2]
- (b) At 5800 K, the temperature of the Sun's surface, many atoms have kinetic energies greater than 1.0 eV. Explain why this is possible even though your answer to (a) is higher than 5800 K. [1]
- (c) State the temperature at which the mean kinetic energy would be 0.50 eV. [1]
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- (a) 7730 K
(3/2)kT = 1.60 × 10⁻¹⁹ J (C1); T = 7730 K (A1)
- (b) (3/2)kT is only the mean; the particles have a range of energies and some have much more than the mean.
B1
- (c) 3860 K
B1 ⟨Eₖ⟩ ∝ T, so half the temperature
- (a) A molecule of mass m moves with velocity component cₓ perpendicular to one face of a cube of side L. It collides elastically with this face. State the change in momentum of the molecule. [1]
- (b) Show that the average force exerted on this face by the molecule is mcₓ²/L. [2]
- (c) The cube contains N identical molecules moving at random. Use your answer to (b) to derive the equation pV = ⅓Nm⟨c²⟩. [3]
- (d) State one point in the derivation where an assumption of the kinetic theory is used, and name the assumption. [1]
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- (a) 2mcₓ (directed away from the wall)
B1 2mcₓ
- (b) Time between collisions with the same face = 2L/cₓ; F = Δp/Δt = 2mcₓ/(2L/cₓ) = mcₓ²/L.
M1 time between collisions 2L/cₓ; A1 force = rate of change of momentum = mcₓ²/L
- (c) Total force = Nm⟨cₓ²⟩/L, so p = F/L² = Nm⟨cₓ²⟩/L³ = Nm⟨cₓ²⟩/V. For random motion ⟨c²⟩ = ⟨cₓ²⟩ + ⟨c_y²⟩ + ⟨c_z²⟩ = 3⟨cₓ²⟩, so pV = ⅓Nm⟨c²⟩.
M1 total force = Nm⟨cₓ²⟩/L and p = F/L²; M1 V = L³; A1 ⟨cₓ²⟩ = ⅓⟨c²⟩ because motion is random (no preferred direction), giving pV = ⅓Nm⟨c²⟩
- (d) e.g. the momentum change 2mcₓ uses perfectly elastic collisions; or ignoring collisions between molecules on the way across uses negligible molecular volume / no intermolecular forces.
B1 any correctly linked assumption
Take the Boltzmann constant k = 1.38 × 10⁻²³ J K⁻¹ and 1 u = 1.66 × 10⁻²⁷ kg.
- (a) Hydrogen molecules have a mass of 2.0 u and oxygen molecules a mass of 32 u. Show that, at the same temperature, the r.m.s. speed of hydrogen molecules is four times that of oxygen molecules. [2]
- (b) In the upper atmosphere the temperature can reach 1000 K. Calculate the r.m.s. speed of hydrogen molecules at this temperature. [2]
- (c) The escape speed from the Earth is 11.2 km s⁻¹. Your answer to (b) is only about one third of this. Suggest why hydrogen has nevertheless escaped from the Earth's atmosphere over its history, but oxygen has not. [2]
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- (a) ½m⟨c²⟩ = (3/2)kT is the same for both, so c_rms ∝ 1/√m; ratio = √(32/2.0) = 4.
M1 same mean KE at same T so c_rms ∝ m^(−1/2); A1 √16 = 4
- (b) 3530 m s⁻¹
c_rms = √(3kT/m) (C1) = √(3 × 1.38 × 10⁻²³ × 1000/(2.0 × 1.66 × 10⁻²⁷)) = 3530 m s⁻¹ (A1)
- (c) The r.m.s. is an average: some molecules move much faster than c_rms, and over millions of years enough reach escape speed and leave; oxygen molecules are four times slower, so the fraction reaching 11.2 km s⁻¹ is negligible.
B1 molecules have a range (distribution) of speeds, some ≫ c_rms; B1 oxygen's c_rms is 4 times smaller so (almost) none reach escape speed
An oxygen molecule of mass 5.3 × 10⁻²⁶ kg moves backwards and forwards at 450 m s⁻¹ between two opposite walls of a cubical box of side 0.20 m, hitting each wall at right angles. Collisions are elastic and the molecule hits nothing else.
- (a) Calculate the change in momentum of the molecule in one collision with the wall. [1]
- (b) Calculate the time between successive collisions of the molecule with this wall. [1]
- (c) Calculate the average force exerted on the wall by this molecule, and the resulting pressure on the wall. [2]
- (d) The box actually contains 2.0 × 10²³ such molecules at the same r.m.s. speed of 450 m s⁻¹, moving randomly. Determine the pressure in the box. [2]
Show answer
- (a) 4.77 × 10⁻²³ N s
Δp = 2mc = 2 × 5.3 × 10⁻²⁶ × 450 = 4.77 × 10⁻²³ N s (B1)
- (b) 8.89 × 10⁻⁴ s
t = 2L/c = 0.40/450 = 8.89 × 10⁻⁴ s (B1)
- (c) F = 5.37 × 10⁻²⁰ N; p = 1.34 × 10⁻¹⁸ Pa
F = Δp/t = 5.37 × 10⁻²⁰ N (B1); p = F/L² = F/0.040 = 1.34 × 10⁻¹⁸ Pa (B1)
- (d) 89400 Pa
M1 only one-third of the motion is perpendicular to each wall on average: p = ⅓Nm⟨c²⟩/V; A1 ⅓ × 2.0 × 10²³ × 5.3 × 10⁻²⁶ × 450²/0.0080 = 89400 Pa
A tube contains a mixture of helium and argon at a temperature of 290 K. Take the Boltzmann constant k = 1.38 × 10⁻²³ J K⁻¹ and 1 u = 1.66 × 10⁻²⁷ kg.
- (a) Calculate the r.m.s. speed of (i) the helium atoms (4.0 u), (ii) the argon atoms (40 u). [3]
- (b) State and explain how the mean kinetic energies of the two kinds of atom compare. [2]
- (c) The mixture is heated until the r.m.s. speed of the argon atoms equals the r.m.s. speed the helium atoms had at 290 K. Determine the new temperature. [2]
Show answer
- (a) (i) 1340 m s⁻¹ (ii) 425 m s⁻¹
C1 c_rms = √(3kT/m); A1 helium 1340 m s⁻¹; A1 argon 425 m s⁻¹
- (b) They are equal, because the gases are in thermal equilibrium at the same temperature and ⟨Eₖ⟩ = (3/2)kT depends only on T.
B1 equal; B1 same T and ⟨Eₖ⟩ depends only on T
- (c) 2900 K
M1 c_rms² ∝ T/m, need c² × 10 for argon so T × 10; A1 2900 K
Nitrogen gas is stored in a cylinder. k = 1.38 × 10⁻²³ J K⁻¹.
- (a) Calculate the number of molecules per cubic metre of the gas at 1.01 × 10⁵ Pa and 273 K. [2]
- (b) Treating each molecule as a sphere of diameter 3.7 × 10⁻¹⁰ m, estimate the fraction of the volume of the gas occupied by the molecules themselves. [2]
- (c) The gas is compressed to 300 times this pressure at the same temperature. Assuming it still behaves ideally, estimate the new fraction, and explain why the gas would no longer behave as an ideal gas. [3]
Show answer
- (a) 2.68 × 10²⁵ m⁻³
N/V = p/kT (C1) = 1.01 × 10⁵/(1.38 × 10⁻²³ × 273) = 2.68 × 10²⁵ m⁻³ (A1)
- (b) 7.1 × 10⁻⁴
volume of one molecule = 4/3 π (1.85 × 10⁻¹⁰)³ = 2.65 × 10⁻²⁹ m³ (C1); fraction = 2.68 × 10²⁵ × 2.65 × 10⁻²⁹ = 7.1 × 10⁻⁴ (A1)
- (c) about 0.21; the volume of the molecules is no longer negligible, and molecules are close enough for intermolecular forces to matter.
B1 0.21 (300 × answer to (b)); B1 assumption of negligible molecular volume fails; B1 intermolecular forces no longer negligible
- (a) Use the kinetic theory equations to show that equal volumes of any two ideal gases at the same pressure and temperature contain the same number of molecules. [3]
- (b) Two flasks, each of volume 0.0250 m³, contain hydrogen and carbon dioxide respectively at 1.2 × 10⁵ Pa and 310 K. Calculate the number of molecules in each flask. (k = 1.38 × 10⁻²³ J K⁻¹) [2]
- (c) State which flask contains the greater mass of gas and explain why the pressures are equal although the molecules have very different masses. [2]
Show answer
- (a) pV = ⅓Nm⟨c²⟩ = ⅔N × ½m⟨c²⟩ = ⅔N(3/2 kT) = NkT; so N = pV/kT, which depends only on p, V and T, not on the type of molecule.
M1 write pV = ⅔N⟨Eₖ⟩; M1 ⟨Eₖ⟩ = (3/2)kT is the same for any gas at the same T; A1 N = pV/kT, same for both
- (b) 7.01 × 10²³ in each
N = pV/kT (C1) = 1.2 × 10⁵ × 0.0250/(1.38 × 10⁻²³ × 310) = 7.01 × 10²³ (A1)
- (c) Carbon dioxide (heavier molecules). Its molecules move more slowly (lower c_rms), so each has the same mean KE; the product m⟨c²⟩ is the same, so the pressure is the same.
B1 carbon dioxide flask; B1 lighter molecules move faster, m⟨c²⟩ (mean KE) equal so p equal
N_A = 6.02 × 10²³ mol⁻¹.
- (a) State what is meant by an ideal gas. [1]
- (b) Define the Avogadro constant. [1]
- (c) A sample of oxygen gas has a mass of 64 g. The molar mass of oxygen (O₂) is 32 g mol⁻¹. Calculate (i) the amount of gas in moles, (ii) the number of oxygen molecules. [2]
Show answer
- (a) A gas that obeys pV ∝ T (pV = nRT) at all pressures, volumes and temperatures.
B1 pV ∝ T (T thermodynamic temperature) for all p, V, T
- (b) The number of particles (atoms or molecules) in one mole of a substance, 6.02 × 10²³ mol⁻¹.
B1 number of particles in one mole (the mole is defined by this fixed number)
- (c) (i) 2.0 mol (ii) 1.20 × 10²⁴
B1 64/32 = 2.0 mol; B1 2.0 × 6.02 × 10²³ = 1.20 × 10²⁴
An oxygen cylinder at a hospital in Lahore has an internal volume of 0.010 m³. The oxygen is at a pressure of 1.37 × 10⁷ Pa and a temperature of 20 °C (293 K). Take R = 8.31 J K⁻¹ mol⁻¹, N_A = 6.02 × 10²³ mol⁻¹ and k = 1.38 × 10⁻²³ J K⁻¹.
- (a) Calculate the amount of oxygen, in mol, in the cylinder. [2]
- (b) Calculate the mass of oxygen in the cylinder. The molar mass of oxygen is 0.032 kg mol⁻¹. [1]
- (c) Use pV = NkT to calculate the number of molecules in the cylinder. [2]
Show answer
- (a) 56.3 mol
n = pV/RT (C1) = 1.37 × 10⁷ × 0.010/(8.31 × 293) = 56.3 mol (A1)
- (b) 1.80 kg
56.3 × 0.032 = 1.80 kg (B1)
- (c) 3.39 × 10²⁵
N = pV/kT (C1) = 1.37 × 10⁷ × 0.010/(1.38 × 10⁻²³ × 293) = 3.39 × 10²⁵ (A1)
Before a journey from Lahore to Islamabad, the air in a car tyre is at a pressure of 2.2 × 10⁵ Pa and a temperature of 15 °C. After the journey the temperature of the air in the tyre is 45 °C.
- (a) Calculate the pressure of the air in the tyre after the journey. Assume the volume of the tyre does not change. [2]
- (b) Explain why the temperatures must be converted to kelvin before using the gas equation. [1]
- (c) State one assumption, apart from constant volume, made in your calculation. [1]
Show answer
- (a) 2.43 × 10⁵ Pa
p₁/T₁ = p₂/T₂ with T in K: 288 K and 318 K (C1); p₂ = 2.2 × 10⁵ × 318/288 = 2.43 × 10⁵ Pa (A1)
- (b) p is proportional to the thermodynamic (absolute) temperature, which is zero when p would be zero; the Celsius scale has an arbitrary zero.
B1
- (c) No air leaks out (amount of gas constant) / air behaves as an ideal gas.
B1
R = 8.31 J K⁻¹ mol⁻¹ and N_A = 6.02 × 10²³ mol⁻¹.
- (a) Show that k = R/N_A and calculate its value. [2]
- (b) Calculate the number of gas molecules in 1.0 cm³ of air at a pressure of 1.0 × 10⁵ Pa and a temperature of 300 K. [2]
Show answer
- (a) k = 8.31/6.02 × 10²³ = 1.38 × 10⁻²³ J K⁻¹
M1 pV = nRT = NkT with N = nN_A, so k = R/N_A; A1 1.38 × 10⁻²³ J K⁻¹
- (b) 2.42 × 10¹⁹
V = 1.0 × 10⁻⁶ m³ (C1); N = pV/kT = 1.0 × 10⁵ × 1.0 × 10⁻⁶/(1.38 × 10⁻²³ × 300) = 2.42 × 10¹⁹ (A1)
A weather balloon is filled with helium at ground level, where the pressure is 1.0 × 10⁵ Pa and the temperature is 290 K. Its volume is 4.0 m³. It rises to a height where the pressure is 3.0 × 10⁴ Pa and the temperature is 240 K.
- (a) Calculate the volume of the balloon at the high altitude. Assume no gas escapes and the pressure inside equals the pressure outside. [2]
- (b) Calculate the amount of helium in the balloon, in mol. (R = 8.31 J K⁻¹ mol⁻¹) [2]
Show answer
- (a) 11.0 m³
p₁V₁/T₁ = p₂V₂/T₂ (C1); V₂ = 4.0 × (1.0 × 10⁵/3.0 × 10⁴) × (240/290) = 11.0 m³ (A1)
- (b) 166 mol
n = pV/RT (C1) = 1.0 × 10⁵ × 4.0/(8.31 × 290) = 166 mol (A1)
A room in Quetta measures 5.0 m × 4.0 m × 3.0 m. Because the city is at high altitude, the air pressure is 8.0 × 10⁴ Pa. The air temperature is 283 K. Take R = 8.31 J K⁻¹ mol⁻¹, N_A = 6.02 × 10²³ mol⁻¹ and k = 1.38 × 10⁻²³ J K⁻¹.
- (a) Calculate the number of air molecules in the room. [2]
- (b) Calculate the mass of air in the room. The mean molar mass of air is 0.029 kg mol⁻¹. [2]
- (c) The heater raises the air temperature to 293 K while the pressure stays constant. Explain what happens to the number of molecules in the room. [2]
Show answer
- (a) 1.23 × 10²⁷
V = 60 m³ (C1); N = pV/kT = 8.0 × 10⁴ × 60/(1.38 × 10⁻²³ × 283) = 1.23 × 10²⁷ (A1)
- (b) 59.2 kg
n = N/N_A = 2040 mol (C1); mass = 2040 × 0.029 = 59.2 kg (A1)
- (c) It decreases (to 1.19 × 10²⁷): N = pV/kT, so with p and V fixed N ∝ 1/T; some air leaves the room through gaps.
B1 N decreases; B1 N ∝ 1/T at constant p and V, air escapes through gaps/windows
Container A, of volume 2.0 × 10⁻³ m³, contains an ideal gas at 3.0 × 10⁵ Pa. Container B, of volume 3.0 × 10⁻³ m³, contains the same gas at 1.0 × 10⁵ Pa. They are joined by a narrow tube with a closed tap. Both containers are at the same constant temperature. The tap is opened.
- (a) Explain why the total value of pV for the two containers is the same before and after the tap is opened. [1]
- (b) Calculate the final pressure. [2]
- (c) Determine the fraction of the total gas that was originally in container A. [1]
Show answer
- (a) pV = nRT; the total amount n and the temperature are unchanged, so Σ pV is unchanged.
B1
- (b) 1.80 × 10⁵ Pa
3.0 × 10⁵ × 2.0 + 1.0 × 10⁵ × 3.0 = p × 5.0 (C1); p = 1.80 × 10⁵ Pa (A1)
- (c) 0.67
B1 6.0/9.0 = 0.67 (ratio of pV values)
A student compresses air in a sealed syringe connected to a pressure gauge, keeping the temperature at 295 K. She records the volume V and pressure p.V/cm³: 20, 25, 30, 40, 50p/kPa: 200.5, 160.2, 133.6, 100.3, 80.1R = 8.31 J K⁻¹ mol⁻¹.
- (a) Explain why a graph of p against 1/V should be a straight line through the origin, and give an expression for its gradient. [2]
- (b) Calculate 1/V for each result, in units of 10⁴ m⁻³. [1]
- (c) Determine the gradient of the graph, using the first and last points, and hence the amount of gas in mol. [3]
- (d) State why the syringe is pushed in slowly. [1]
Show answer
- (a) pV = nRT so p = nRT × (1/V): linear with zero intercept for constant n and T; gradient = nRT.
B1 p ∝ 1/V at constant T and n; B1 gradient = nRT
- (b) 1/V (×10⁴ m⁻³): 5.00, 4.00, 3.33, 2.50, 2.00
B1 all values (V in m³: 20 cm³ = 2.0 × 10⁻⁵ m³ → 5.00 × 10⁴ m⁻³)
- (c) gradient = 4.01 J (N m); n = 1.64 × 10⁻³ mol
C1 gradient = (200.5 − 80.1) × 10³/((5.00 − 2.00) × 10⁴) = 4.01 J; C1 n = gradient/RT; A1 n = 4.01/(8.31 × 295) = 1.64 × 10⁻³ mol
- (d) So that the temperature of the gas stays constant (work done on the gas heats it if compressed quickly).
B1
R = 8.31 J K⁻¹ mol⁻¹.
- (a) Show that the density ρ of an ideal gas of molar mass M is given by ρ = pM/(RT). [2]
- (b) Calculate the density of carbon dioxide (M = 0.044 kg mol⁻¹) and of air (M = 0.029 kg mol⁻¹) at 1.01 × 10⁵ Pa and 298 K. [2]
- (c) Use your answers to explain why carbon dioxide leaking from a fire extinguisher collects at floor level. [1]
Show answer
- (a) n = m/M, so pV = (m/M)RT and ρ = m/V = pM/(RT).
M1 n = mass/M; A1 rearrangement to ρ = pM/RT
- (b) CO₂: 1.79 kg m⁻³; air: 1.18 kg m⁻³
B1 0.044 × 1.01 × 10⁵/(8.31 × 298) = 1.79; B1 0.029 × 1.01 × 10⁵/(8.31 × 298) = 1.18
- (c) At the same p and T, CO₂ is about 1.5 times denser than air, so it sinks below the air.
B1
Two glass bulbs X and Y, each of volume 1.5 × 10⁻³ m³, are joined by a thin tube of negligible volume. Initially bulb X is at 300 K and bulb Y at 400 K, and the gas pressure throughout is 1.0 × 10⁵ Pa. R = 8.31 J K⁻¹ mol⁻¹.
- (a) Calculate the total amount of gas in the two bulbs, in mol. [2]
- (b) Bulb Y is now heated to 500 K, while bulb X is kept at 300 K. Gas flows through the tube until the pressures are equal. Calculate the new pressure. [3]
- (c) State the direction in which gas flows while bulb Y is being heated, and explain why. [1]
Show answer
- (a) 0.105 mol
n = pV/RT for each bulb: 1.0 × 10⁵ × 1.5 × 10⁻³/(8.31 × 300) + 1.0 × 10⁵ × 1.5 × 10⁻³/(8.31 × 400) (C1) = 0.0602 + 0.0451 = 0.105 mol (A1)
- (b) 1.09 × 10⁵ Pa
total n conserved: (pV/R)(1/300 + 1/500) = 0.105 (C1)(M1); p = 1.0 × 10⁵ × (1/300 + 1/400)/(1/300 + 1/500) = 1.09 × 10⁵ Pa (A1)
- (c) From Y to X: heating Y raises its pressure above that in X, so gas moves to X until pressures are equal.
B1
A student investigates how the volume V of a fixed mass of air, trapped in a narrow glass tube by a thread of oil, varies with temperature θ at constant pressure. The tube is placed in a water bath, which is stirred and heated.θ/°C: 10, 30, 50, 70, 90V/cm³: 28.3, 30.4, 32.3, 34.3, 36.3
- (a) Explain why a graph of V against θ is expected to be a straight line, and what the intercept on the θ-axis represents. [2]
- (b) Determine the gradient of the line, using the first and last results as points on the best-fit line. [1]
- (c) Use your gradient to determine the value of absolute zero in °C. [2]
- (d) The pressure of the trapped air is 1.01 × 10⁵ Pa. Use the result at 10 °C to calculate the amount of air trapped, in mol. (R = 8.31 J K⁻¹ mol⁻¹) [2]
- (e) Suggest two reasons why the extrapolation to V = 0 may not give an accurate value for absolute zero. [2]
Show answer
- (a) At constant p, V = (nR/p)T = (nR/p)(θ + 273), which is linear in θ; V = 0 where θ = −273 °C, so the θ-intercept represents absolute zero.
B1 V ∝ T at constant p, linear in θ; B1 intercept (V = 0) gives absolute zero
- (b) 0.100 cm³ K⁻¹
B1 (36.3 − 28.3)/(90 − 10) = 0.100 cm³ K⁻¹ (accept °C⁻¹)
- (c) −273 °C
M1 θ-intercept = 10 − 28.3/0.100; A1 −273 °C
- (d) 1.22 × 10⁻³ mol
n = pV/RT with V = 28.3 × 10⁻⁶ m³ and T = 283 K (C1); n = 1.22 × 10⁻³ mol (A1)
- (e) Long extrapolation magnifies small errors in the gradient; the thread/trapped air may not be at the bath temperature (insufficient time); water vapour in the trapped air; real gas liquefies before reaching absolute zero.
B1 each for any two sensible reasons
A steel cylinder of fixed volume contains 2.4 kg of compressed gas at 6.0 × 10⁵ Pa and 300 K. Over a cold night the valve leaks. In the morning the pressure is 4.2 × 10⁵ Pa and the temperature is 290 K.
- (a) Determine the fraction of the original gas that remains in the cylinder. [3]
- (b) Calculate the mass of gas that has escaped. [1]
- (c) A student calculates the fraction remaining as 4.2/6.0 = 0.70. Explain why this is incorrect. [1]
Show answer
- (a) 0.724
C1 n ∝ p/T at constant V; C1 n₂/n₁ = (4.2 × 10⁵/290)/(6.0 × 10⁵/300); A1 0.724
- (b) 0.662 kg
B1 remaining mass = 2.4 × 0.724 = 1.74 kg, so 0.662 kg escaped
- (c) The temperature has also fallen; at the lower temperature the same amount of gas would have a lower pressure, so p alone does not measure the amount.
B1
